is a weak tribasic acid with , and . The value of pX of solution, where and , is — Ionic Equilibrium Chemistry Question
Question
$\text{H}_3\text{A}$ is a weak tribasic acid with $K_{a1} = 10^{-5}$, $K_{a2} = 10^{-9}$ and $K_{a3} = 10^{-13}$. The value of pX of $0.1 \text{ M} – \text{H}_3\text{A}$ solution, where $\text{pX} = -\log_{10}\text{X}$ and $\text{X} = \frac{[\text{A}^{3-}]}{[\text{HA}^{2-}]}$, is
💡 Solution & Explanation
The ratio $\text{X}$ is derived from the third dissociation step: $\text{HA}^{2-} \rightleftharpoons \text{H}^+ + \text{A}^{3-}$, where $K_{a3} = \frac{[\text{H}^+][\text{A}^{3-}]}{[\text{HA}^{2-}]} = [\text{H}^+] \text{X}$. Thus, $\text{X} = \frac{K_{a3}}{[\text{H}^+]}$. $[\text{H}^+]$ is defined by the first dissociation: $[\text{H}^+] \approx \sqrt{K_{a1} C} = \sqrt{10^{-5} \times 0.1} = 10^{-3} \text{ M}$. Substituting the values gives $\text{X} = \frac{10^{-13}}{10^{-3}} = 10^{-10}$. Therefore, $\text{pX} = -\log(10^{-10}) = 10.0$. Therefore, correct answer is D.